小编pre*_*ash的帖子

FAILURE: Build failed with an exception.

* What went wrong:
A problem occurred configuring root project 'oscar'.
> Could not open proj class cache for build file '/Users/prakashv/dev/workspace/oscar-api/build.gradle' (/Users/prakashv/.gradle/caches/2.6/scripts/build_7o20dvaacd2xu3ci3o56uhd4f/proj).
   > Timeout waiting to lock proj class cache for build file '/Users/prakashv/dev/workspace/oscar-api/build.gradle' (/Users/prakashv/.gradle/caches/2.6/scripts/build_7o20dvaacd2xu3ci3o56uhd4f/proj). It is currently in use by another Gradle instance.
     Owner PID: unknown
     Our PID: 47991
     Owner Operation: unknown
     Our operation: Initialize cache
     Lock file: /Users/prakashv/.gradle/caches/2.6/scripts/build_7o20dvaacd2xu3ci3o56uhd4f/proj/cache.properties.lock

* Try:
Run with --stacktrace option to get the stack trace. Run with …

CompletableFuture::supplyAsync(() -> IO bound queries)

如何为CompletableFuture :: supplyAsync选择Executor以避免污染ForkJoinPool.commonPool().

有许多选项Executors(newCachedThreadPool,newWorkStealingPool,newFixedThreadPool等)

我在这里阅读了关于新ForkJoinPool的内容

如何为我的用例选择合适的？

我试图从其他客户端测试经过身份验证的API网关端点.如何在发出请求时生成/设置"AWS_IAM"授权标头？

我得到了一个不想要的'？' 在执行此操作时,在我的文件名末尾:

emplid=$(grep -a "Student ID" "$i".txt  | sed 's/(Student ID:  //g' | sed 's/)Tj//g' ) 
 #gets emplid by doing a grep from some text file
echo "$emplid"   #prints employee id correctly 
cp "$i" "$emplid".pdf  #getting an extra '?' character after emplid and before .pdf

即不是得到像123456.pdf这样的文件名,而是得到123456？.pdf.如果回波打印正确,为什么会发生这种情况？如何删除尾随问号字符？

是否有内置函数可以在C/C++中的2个给定限制之间生成随机素数？

我想要一个可以生成100万到10亿之间随机素数的函数

是否可以构建二进制搜索树只给出其前序遍历？

我知道只有在给定inorder和preorder遍历的情况下才能构造二叉树.但我的问题是二元搜索树特有的.

例如:鉴于:5,3,1,4,7,8

  Construct : 

       5
    3    7 
  1   4    8

这是我的模板

{  
   "AWSTemplateFormatVersion":"2010-09-09",
   "Resources":{  
      "lambdafunction":{  
         "Type":"AWS::Lambda::Function",
         "Properties":{  
            "Handler":"index.handler",
            "Role":{  
               "Fn::GetAtt":[  
                  "RootRole",
                  "Arn"
               ]
            },
            "Code":{  
               "S3Bucket":"{s3_bucket_name}",
               "S3Key":"lambda-zip"
            },
            "Runtime":"java8",
            "Timeout":"25"
         }
      },
      "RootRole":{  
         "Type":"AWS::IAM::Role",
         "Properties":{  
            "AssumeRolePolicyDocument":{  
               "Version":"2012-10-17",
               "Statement":[  
                  {  
                     "Effect":"Allow",
                     "Principal":{  
                        "Service":[  
                           "ec2.amazonaws.com"
                        ]
                     },
                     "Action":[  
                        "sts:AssumeRole"
                     ]
                  }
               ]
            },
            "Path":"/",
            "Policies":[  
               {  
                  "PolicyName":"root",
                  "PolicyDocument":{  
                     "Version":"2012-10-17",
                     "Statement":[  
                        {  
                           "Effect":"Allow",
                           "Action":"*",
                           "Resource":"*"
                        }
                     ]
                  }
               }
            ]
         }
      }
   }
}

堆栈创建后lambda函数的名称是lambda-lambdafunction-18SJKJ5Q40AKZ IAM角色的名称是lambda-RootRole-12S8E9CA0EOVM

该模板似乎没有办法定义lambda函数名称http://docs.aws.amazon.com/AWSCloudFormation/latest/UserGuide/aws-resource-lambda-function.html.而且我不确定为什么最后会附加随机字符.

我有一个异步执行的查询输入流。我想确保当我使用时Completablefuture::join，这些要求的结果是按照输入查询流的顺序收集的。

这是我的代码的样子：

queries.stream()
     .map(query -> CompletableFuture.supplyAsync(() -> {
                    try {
                        return SQLQueryEngine.execute(query);
                    } catch (InternalErrorException e) {
                        throw new RuntimeException(e);
                    }
     }))
     .map(CompletableFuture::join)
     .collect(Collectors.toList());

SQLQueryEngine.execute(查询); 返回一个List<Results>so 输出是List<List<Result>。我想展平并将所有结果合并到一个列表中。如果我在收集之前使用 .flatMap(List::stream) 来扁平化，它会保持排序吗？